Two satellites of mass m and 9m are orbiting | vedclass

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(C) According to Kepler's third law of planetary motion,the square of the time period $T$ of a satellite is proportional to the cube of the orbital ra

Vedclass · Accepted Answer

$1: 1$

Vedclass · Answer

(C) According to Kepler's third law of planetary motion,the square of the time period $T$ of a satellite is proportional to the cube of the orbital radius $R$,given by $T^2 \propto R^3$. This implies $T = 2\pi \sqrt{\frac{R^3}{GM}}$,where $G$ is the gravitational constant and $M$ is the mass of the planet. Note that the time period $T$ depends only on the radius of the orbit $R$ and the mass of the central planet $M$. It is independent of the mass of the satellite $m$. Since both satellites orbit the same planet at the same radius $R$,their time periods will be equal. Therefore,the ratio of their periods of revolution is $1: 1$.

Two satellites of mass $m$ and $9m$ are orbiting a planet in orbits of radius $R$. Their periods of revolution will be in the ratio of

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